Albert Einstein theoretically proposed the existence of gravitational waves in the year 1915.

Ground wave propagation: frequency less than 2 MHz
Sky wave propagation: 3 to 30 MHz
Space wave propagation: Above 30 MHz to 400 GHz
Satellite communication: uplink communication 6 GHz band downlink communication 4 GHz band

K.E= hv - Φ
K.E₁ = 0.9 - 0.6 = 0.3 eV
K.E₂ = 3.3 - 0.6 = 2.7 ev
K.E ∝ v²
\(\frac{0.3}{2.7}=\frac{v^2_1}{v^2_2} \)
\(\frac{v^1}{v^2} =\frac{1}{3}\)

2n t cos r = (2m + 1) \(\frac{\lambda}{2}\)
For maximum
m =  2 (For visible region), n - refractive index.
cos r = cos 0 = 1
t = 5 x 10^-5 x 10^-2 = 5 x 10^-7 m
\(n=\frac{(2m+1)\frac{\lambda}{2}}{2t}=\frac{5\lambda}{2 \times 2 \times t}\)
\(=\frac{5 \times5320\times10^{-10}}{4 \times 5 \times 10^{-7}}\)
\(=\frac{5 \times5320\times10^{-10}}{20}=1330 \times 10^3\)
n = 1.330

Total power = 15 x 40 + 5 x 100 + 5 x 80 + 1000
= 600 + 500 + 400 + 1000
= 2500 W
P = VI, V = 220 V
\(I=\frac{P}{V}=\frac{2500}{220}=11.363 \ A\)
≃ 12 A

(i) The satellite communication is a mode of transmission of signal between transmitter and receiver via satellite.
(ii) The message signal from the Earth station is transmitted to the satellite on board via an uplink (frequency band 6 GHz), amplified by a transponder and then retransmitted to another earth station via a downlink (frequency band 4 GHz).
Applications:
Satellites are classified into different types based on their applications.
(i) Weather Satellites:
They are used to monitor the weather and climate of Earth. By measuring cloud mass, these satellites enable us to predict rain and dangerous storms like hurricanes, cyclones etc.
(ii) Communication satellites:
They are used to transmit television, radio, internet signals etc. Multiple satellites are used for long distance communication
(iii) Navigation satellites:
These are employed to determine the geographic location of ships, aircraft or any other object.

Advantages of robotics:
(i) The robots are much cheaper than humans.
(ii) Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
(iii) Robots are more precise and error free in performing the task.
(iv) Stronger and faster than humans.
(v) Robots can work in extreme environmental conditions: extreme hot or cold, space or underwater. In dangerous situations like bomb detection and bomb deactivation.
(vi) In warfare, robots can save human lives.
(vii) Robots are significantly used in handling materials in chemical industries especially in nuclear plants which can lead to health hazards in humans.
Disadvantages of Robotics:
(i) Robots have no sense of emotions or conscience.
(ii) They lack empathy and hence create an emotionless workplace.
(iii) If ultimately robots would do all the work, and the humans will just sit and monitor them, health hazards will increase rapidly.
(iv) Unemployment problem will increase.
(v) Robots can perform defined tasks and cannot handle unexpected situations.
(vi) The robots are well programmed to do a job and if a small thing goes wrong it ends up in a big loss to the company.
(vii) If a robot malfunctions, it takes time to identify the problem, rectify it, and even reprogram if necessary. This process requires signi cant time.
(viii) Humans cannot be replaced by robots in decision making.
(ix) Till the robot reaches the level of human intelligence, the humans in work place will exit.

The initial number of molecules, N_o = 150
The final number of molecules, N = 150
\(\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n } \ or \ \frac { 150 }{ 600 } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }\)
\({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }={ \left( \frac { 1 }{ 2 } \right) }^{ n } \ or \ n=2=\frac { t }{ { T }_{ 1/2 } } \)
t = 2 x 600 s = 1200 s
Best of disintegration,
\(R=\frac { dN }{ dt } =-\lambda N\)
\(=\frac { 0.693 }{ { T }_{ 1/2 } } \times 150\)
\(=\frac { 0.693 }{ { 600} } \times 150\) = 0.173
disintegration/second at the instant when 150 molecules were remaining.

E and B vectors must be in x and y directions.
Formula: We know \(\lambda =\frac { v }{ \gamma } =\frac { 3\times { 10 }^{ 8 } }{ 30\times { 10 }^{ 6 } } \)
λ = 10m.

I₁ = 4A, I₂= 6A, r = 3cm = 0.03A
\(\frac { F }{ l } =\frac { { \mu }_{ 0 } }{ 4\pi } \frac { 2{ I }_{ 1 }{ I }_{ 2 } }{ 0.03 } \)
\(\frac { F }{ 1 } =\frac { 10^{ -7 }\times 2\times 4\times 6 }{ 0.03 } \)
F = 1.6 x 10^-4 N/m.

Given:
Induced emf, e = Blv (or) e = \(\frac12\) Bl²ω [where, v = rω, 1 = 2R]
Also, ω=2ㅠf = 2π x \(\frac{120}{60}\)
ω = 4ㅠ
Solution:
emf, e = \(\frac12\) x 4ㅠ x 0.4 x 10^-4 (0.5)²
= 6.28 x 10^-5V
The number of spokes is immaterial because the emf's across the spokes are in parallel

(i) Galvanometer to an Ammeter:

(i) Ammeter is an instrument used to measure current flowing in the electrical circuit.
(ii) The Ammeter must offer low resistance such that it will not change the current passing through it. So, ammeter is connected in series to measure the circuit current.
(iii) A galvanometer is converted into an ammeter by connecting a low resistance in parallel with the galvanometer.
(iv) Let I be the current passing through the circuit. When current I reaches the junction A, it divides into two components.
a) I_g → Current passing through the galvanometer
b) I - I_g → Current passing through the shunt resistance.
(v) The potential difference across the galvanometer is same as the potential difference across the shunt resistance.
\(\mathrm{V}_{\text {galvanometer }} =\mathrm{V}_{\text {shunt }} \)
\(\Rightarrow \mathrm{I}_{\mathrm{g}} \mathrm{R}_{\mathrm{g}} =\left(\mathrm{I}-\mathrm{I}_{g}\right) \mathrm{S} \)
\(\mathrm{S} =\frac{I_{g}}{\left(I-I_{g}\right)} R_{g} \)  (or)
\(\mathrm{I}_{\mathrm{g}}=\frac{S}{S+R_{g}} I \Rightarrow I_{g} \propto I\)
Since, the deflection in the galvanometer is proportional to the current passing through it.
\(\theta=\frac{1}{G} I_{g} \Rightarrow \theta \propto I_{g} \Rightarrow \theta \propto I\)
Where, R_g → Galvanometer resistance, S → Shunt resistance.
Since shunt resistance is connected in parallel to galvanometer,
Effective resistance,\(\frac{1}{R_{e f f}}=\frac{1}{R_{g}}+\frac{1}{S} \Rightarrow R_{e f f}=\frac{R_{g} S}{R_{g}+S}=R_{a}\)
R_a ⇒ low resistance. An ideal ammeter has zero resistance.
The percentage error in measuring a current through an ammeter is,
\(\frac{\Delta I}{I} \times 100 \%=\frac{I_{i d e a l}-I_{a c t u a l}}{I_{a c t u a l}} \times 100 \%\)
(ii) Galvanometer to a voltmeter:
i) A voltmeter is an instrument used to measure potential difference across any two points in the electrical circuits. 
ii) Voltmeter must have high resistance and when it is connected in parallel, it will rot draw appreciable current so that it will indicate the true potential difference.
iii) A galvanometer is converted into a voltmeter by connecting high resistance  R_h in series with galvanometer.
iv) Let R_g be the resistance of galvanometer and I_g be the current with which the galvanometer produces full scale deflection.
v) Since the galvanometer is connected in series with high resistance, the current in the electrical circuit is same as the current passing through the galvanometer.

\(\mathrm{I}=\mathrm{I}_{\mathrm{g}} \)
\(\mathrm{I}=I_{g} \Rightarrow I_{g}=\frac{\text { potential difference }}{\text { total resistance }} \)
Since the galvanometer and high resistance are connected in series, the voltmeter resistance is,
\(R_{v} =R_{g}+R_{h} \)
Therefore,
\(I_{g} =\frac{V}{R_{g}+R_{h}} \)
\(\Rightarrow R_{h} =\frac{V}{I_{g}}-R_{g} \)
Note that \(I_{g} \propto V\)
R_h is very large. An ideal voltmeter has infinite resistance

Conservation of energy:
(i) The truth of Lenz's law can be established on the basis of the law of conservation of energy. The explanation is as follows:
(ii) According to Lenz's law, when a magnet is moved either towards or away from a coil, the induced current produced opposes its motion
(iii) As a result, there will always be a resisting force on the moving magnet.
(iv) Work has to be done by some external agency to move the magnet against this resisting force
(v) Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into Joule heat in the coil i.e., energy is converted from one form to another.
(vi) On the contrary to Lenz's law, let us assume that the induced current helps the cause responsible for its production.
(vii) Now When we push the magnet litle bit towards the coil, the induced current helps the movement of the magnet towards the coil.
(viii) Then the magnet starts moving towards the coil without any expense of energy. This, becomes a perpetual motion machine.
(ix) In practice, no such machine is possible. Therefore, the assumption that the induced current helps the cause is wrong.

(i) The meter bridge is another form of Wheatstone's bridge. It consists of a uniform manganin wire AB of one meter length.
(ii) This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G₁ and G₂.
(iii) An unknown resistance P is connected in G₁ and a standard resistance Q is connected in G_2. A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR).
(iv) The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.

(v) The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the position of jockey at the wire be at J.
(vi) The resistances corresponding to AJ and JB of the bridge wire now form the resistance R and S of the Wheatstone's bridge. Then for the bridge balance.
\(\cfrac { P }{ Q } =\cfrac { R }{ S } =\cfrac { { r }.AJ }{ { r }.JB } \)   
where r' is the resistance per unit length of wire
\(\cfrac { P }{ Q } =\cfrac { AJ }{ JB } =\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \)   
\(P=Q\cfrac { { l }_{ 1 } }{ { l }_{ 2 } } \)
(vii) By interchanging P and Q, another set of readings are taken and the average value of P is value of unknown resistance.   

The angular resolution has a unit in radian (rad) and it is given by the equation,
\(\theta =\cfrac { 1.22\lambda }{ \alpha } \)

Manipulation of colours by adjusting the size of nano particles with which the materials are made.

The undesired electrical signals are termed as noise. When a signal is transmitted, the undesired signals get mixed with it, leading to distortion of the signal.

The cut-off wavelength of the characteristic x - rays is
\({ \lambda }_{ ° }\frac { 12400 }{ V } \mathring { A } =\frac { 12400 }{ 20000 } \mathring { A }\)
= 0.62 \(\mathring { A } \)
The corresponding frequency is
\({ \upsilon }_{ o }=\frac { c }{ { \lambda }_{ o } } =\frac { 3\times 10^{ 8 } }{ 0.62\times 10^{ -10 } }\) = 4.84 x 10¹⁸ Hz

The moderator is a material used to convert fast neutrons into slow neutrons
(i) A billiard ball striking a stationary billiard ball of equal mass would itself be stopped but the same billiard ball bounces off almost with same speed when it strikes a heavier mass.
(ii) This is the reason for using lighter nuclei as moderators.

The de Broglie wavelength associated with the kinetic energy k is given as \(\lambda=\frac{h}{\sqrt{2 m k}}\) , where m is the mass of the particle.
Therefore \(\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{k}_{\mathrm{e}}}} \text { and } \lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{k}_{\alpha}}} \text {. But } \mathrm{k}_{\mathrm{e}}=\mathrm{k}_{\alpha} . \)
Therefore \(\frac{\lambda_{\mathrm{e}}}{\lambda_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\mathrm{e}}}} \mathrm{m}_{\alpha}>\mathrm{m}_{\mathrm{e}^{*}} \text {. Therefore, } \lambda_{\mathrm{e}}>\lambda_{\alpha^{\circ}}\)

The output at the 1^st AND gate : A\(\overline { B } \)
The output at the 2^nd AND gate : ĀB
The output at the OR gate: Y = A. \(\overline { B } \) + Ā .B

(i) If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases
(ii) It is necessary for sustained oscillations.

UV light is absorbed by the ozone layer
(i) Source: Sun, arc and ionized gases.
(ii) Uses: To destroy bacteria, sterilizing the surgical instruments, burglar alarm etc

(i) In electric field line is the path of movement of a positive test charge (q₀ ⇾ 0) A moving charge experiences a continuous force is an electric field, so field line is always continuous
(ii) The field lines nevel intersect since if they cross, there will be two directions of electric field at the point of intersection, which is impossible.

Power factor is defined as the ratio of resistance to the impedance of an AC circuit
Power factor \(\cos \phi=\frac{R}{2}\)
\(=\frac{Resistance}{Impedance}\)

Balancing lengths are
1₁ = 35 cm, l₂ = 63 cm
E₁ = 1.25v E₂ = ?
At the following point \(E \propto 1\)
\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}} \)
\(\therefore \frac{1.25}{E_{2}}=\frac{35}{63} \quad \therefore E_{2}=\frac{1.25 \times 63}{35}=2.25 \mathrm{v} \)
∴ EMF of the second cell = 2.25 V
V_d = 0.03 x 10^-3 m s^-1

Metal A
Since work function W = hv_o
and v_o > V_o so work function of metal A is more.
Aliter :
On stopping potential axis \(-\frac { { W }_{ o }^{ ' } }{ e } >-\frac { { W }_{ o } }{ e } \)
Hence work function W'o of metal A is more.

(i) The phenomenon of spontaneous emission of highly penetrating radiations such as α, β, γ rays by heavy elements having atomic number greater than 82 is called radioactivity and the substances which emit these radiations are called radioactive elements.
(ii) The radioactive phenomenon spontaneous and is unaffected by any external agent like temperature, pressure, and magnetic fields.

Given, focal length of first lens, f₁ = –70 cm,
focal length of second lens, f₂ = 150 cm.
Equation for focal length of lenses in contact, \(\cfrac { 1 }{ f } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } \) 
Substituting the values,
\(\cfrac { 1 }{ f } =\cfrac { 1 }{ -70 } +\cfrac { 1 }{ 150 } =\cfrac { 1 }{ 70 } +\cfrac { 1 }{ 150 } \) 
\(\cfrac { 1 }{ f } =\cfrac { -150+70 }{ 70\times 150 } =\cfrac { -80 }{ 70\times 150 } =\cfrac { 80 }{ 10500 } \) 
\(f=\cfrac { -1050 }{ 8 } =131.25cm\) 
As the focal length is negative, the combination of two lenses is a diverging system of lenses
The power of the combination is,
\(P=\cfrac { 1 }{ f } =\cfrac { 1 }{ -1.3125m } =0.76D\)

\(\mathrm{m}_{\alpha}=100 ; \mathrm{f}_{o}=0.5 \mathrm{~cm} ; \mathrm{f}_{\mathrm{e}}=? \)
\(\mathrm{~L}_{\alpha}=6.5 \mathrm{~cm}, \mathrm{D}=25 \mathrm{~cm} \)
When the image is formed at infinity,
\(\mathrm{m}_{\alpha}=\frac{\left(\mathrm{L}_{\alpha}-\mathrm{f}_{0}-\mathrm{f}_{\mathrm{c}}\right) \mathrm{D}}{\mathrm{f}_{0} \mathrm{f}_{\mathrm{e}}} \)
\(100 =\left(\frac{6.5-0.5-f_{e}}{0.5 \times f_{e}}\right) \times 25 \)
\(=\left(\frac{6-f_{e}}{0.5 f_{e}}\right) \times 25 \)
\(100 \times 0.5 f_{e} =150-25 f_{e} \)
\(50 f_{e} =150-25 f_{e} \)
\(75 f_{e} =150 \)
\(f_{e} =150 / 75=2 \mathrm{~cm} \)

Conduction current Ie = Displacement current ID
\({ I }_{ C }={ I }_{ s }=\frac { dq }{ dt } =\frac { d }{ dt } ({ q }_{ 0 }cos2\pi \gamma t)\)
\(=-2\pi { q }_{ 0 }\gamma sin2\pi \gamma t\)

The magnetic field due to street wires AB & CD is zero since either θ = 0° or 180° and that due to a semi -circular arc are equal  & opposite. Hence net field at O is zero.

(i) This device consists of a long thick copper rod passing from top of the building to the ground. The upper end of the rod has a sharp spike or a sharp needle as shown in Figure 1.64 (a) and (b).
(ii) The lower end of the rod is connected to the copper plate which is buried deep into the ground. When a negatively charged cloud is passing above the building, it induces a positive charge on the spike.
(iii) Since the induced charge density on thin sharp spike is large, it results in a corona discharge.
(iv) This positive charge ionizes the surrounding air which in turn neutralizes the negative charge in the cloud.

(v) The negative charge pushed to the spikes passes through the copper rod and is safely diverted to the earth.
(vi) The lightning arrester does not stop the lightning; rather it diverts the lightning to the ground safely.